import java.util.LinkedList;
import java.util.Queue;

public class BinaryTree {

    class  TreeNode{
        String val;
        TreeNode left;
        TreeNode right;

        TreeNode(String x){
            val = x;
        }
    }

    public TreeNode createBinaryTree(){
        TreeNode A = new TreeNode("A");
        TreeNode B = new TreeNode("B");
        TreeNode C = new TreeNode("C");
        TreeNode D = new TreeNode("D");
        TreeNode E = new TreeNode("E");
        TreeNode F = new TreeNode("F");
        TreeNode G = new TreeNode("G");
        TreeNode H = new TreeNode("H");

        A.left = B;
        A.right = C;
        B.left = D;
        B.right = E;
        C.left = F;
        C.right = G;
        D.left = H;

        return  A;
    }

    public void preOrder(TreeNode root){
        if(root == null){
            return;
        }
        System.out.print(root.val);
        preOrder(root.left);
        preOrder(root.right);
    }

    public void inOrder(TreeNode root){
        if(root == null){
            return;
        }
        inOrder(root.left);
        System.out.print(root.val);
        inOrder(root.right);
    }

    public void postOrder(TreeNode root){
        if(root == null){
            return;
        }
        postOrder(root.left);
        postOrder(root.right);
        System.out.print(root.val);
    }

    public static int nodeSize;

    /**
     * 获取树中节点的个数：遍历思路
     */
    public int  size(TreeNode root) {
        if(root == null){
            return 0;
        }
        nodeSize++;
        size(root.left);
        size(root.right);
        return nodeSize;
    }

    /**
     * 获取节点的个数：子问题的思路
     *
     * @param root
     * @return
     */
    public int size2(TreeNode root) {
        if(root == null){
            return 0;
        }
        return size2(root.left) + size2(root.right) + 1;
    }


    /*
     获取叶子节点的个数：遍历思路
     */
    public static int leafSize = 0;

    public int  getLeafNodeCount1(TreeNode root) {
        if(root == null){
            return 0;
        }
        if(root.left == null && root.right == null){
            leafSize++;
        }
        getLeafNodeCount1(root.left);
        getLeafNodeCount1(root.right);
        return leafSize;
    }

    /*
     获取叶子节点的个数：子问题
     */
    int getLeafNodeCount2(TreeNode root) {
        if(root == null){
            return 0;
        }
        if(root.left == null && root.right == null){
            return 1;
        }
        return getLeafNodeCount2(root.left) + getLeafNodeCount2(root.right);
    }

    /*
    获取第K层节点的个数
     */
    public int getKLevelNodeCount(TreeNode root, int k) {
        if(root == null){
            return 0;
        }
        if(k == 1) {
            return 1;
        }
        return getKLevelNodeCount(root.left,k-1) + getKLevelNodeCount(root.right,k-1);
    }

    /*
     获取二叉树的高度
     时间复杂度：O(N)
     */
    int getHeight(TreeNode root) {
        if(root == null){
            return 0;
        }
        return Math.max(getHeight(root.left), getHeight(root.right)) + 1;
    }


    // 检测值为value的元素是否存在
    TreeNode find(TreeNode root, String val) {
        if(root == null){
            return null;
        }else {
            if(root.val == val){
                return root;
            }

            TreeNode left = find(root.left, val);
            if(left != null){
                return left;
            }
            TreeNode right = find(root.right, val);
            if(right != null){
                return right;
            }
        }
        return null;
    }

    //层序遍历
    /*void levelOrder(TreeNode root) {

    }*/


    // 判断一棵树是不是完全二叉树
    boolean isCompleteTree(TreeNode root) {
        if(root == null){
            return true;
        }
        Queue<TreeNode> q = new LinkedList<>();
        TreeNode curr = root;
        q.offer(root);

        while(!q.isEmpty()){
            curr = q.poll();
            if(curr!=null){
                q.offer(curr.left);
                q.offer(curr.right);
            }else {
                break;
            }
        }
       while(!q.isEmpty()){
           TreeNode peek = q.poll();
           if(peek != null){
               return false;
           }
       }
        return false;
    }

}
